Gauss's Hypergeometric Theorem/Proof 1

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Theorem

$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$


Proof

Let $x, y, n \in \C$ be complex numbers such that $\map \Re {x + y + n + 1} > 0$.

Let $u \in \C$ be a complex number such that $\cmod u < 1$.

Expanding the product of $\paren {1 + u}^{y + n}$ and $\paren {\dfrac {1 + u} u}^x$:

\(\ds \paren {1 + u}^{y + n}\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \binom {y + n} k u^k\) Binomial Theorem - Complex Numbers
\(\ds \paren {1 + \dfrac 1 u}^x\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \binom x k u^{-k}\) Binomial Theorem - Complex Numbers, Power of Product
\(\ds \leadsto \ \ \) \(\ds \paren {1 + u}^{y + n} \paren {\dfrac {1 + u} u}^x\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \binom {y + n} k u^k \sum_{k \mathop = 0}^\infty \binom x k u^{-k}\) multiplying

The coefficient $a_n$ of $u^n$ of the product $\dfrac {\paren {1 + u}^{x + y + n} } {u^x}$ above can be determined by setting $k = k + n$ in the series with $\dbinom {y + n} k$:

\(\ds a_n\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \binom {y + n} {k + n} \binom x k\) substituting $k = k + n$ in first series: $u^n = u^{k+n} u^{-k}$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {y + n}!} {\paren {k + n}! \paren {y - k}!} \dfrac {x!} {k! \paren {x - k}!}\) Definition of Binomial Coefficient
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {y + n}!} {\paren {k + n}! \paren {y - k}!} \dfrac {x!} {k! \paren {x - k}!} \paren {\dfrac {n! y!} {n! y!} }\) multiplying by $1$
\(\ds \) \(=\) \(\ds \dfrac {\paren {y + n}! } {n! y!} \sum_{k \mathop = 0}^\infty \dfrac {n! } {k! \paren {k + n}! } \dfrac {y!} {\paren {y - k}!} \dfrac {x!} {\paren {x - k}!}\) moving $\dfrac {\paren {y + n}! } {n! y!}$ outside the sum and rearranging
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} } \sum_{k \mathop = 0}^\infty \dfrac {n! } {k! \paren {k + n}! } \dfrac {y!} {\paren {y - k}!} \dfrac {k!} {k!} \dfrac {x!} {\paren {x - k}!} \dfrac {k!} {k!}\) multiplying by $1$
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} } \sum_{k \mathop = 0}^\infty \dfrac {n! } {k! \paren {k + n}! } \dbinom y k \dbinom x k \paren {k!}^2\) Definition of Binomial Coefficient
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} } \sum_{k \mathop = 0}^\infty \dfrac {n! } {k! \paren {k + n}! } \paren {-1}^k \dbinom {-y + k - 1} k \paren {-1}^k \dbinom {-x + k - 1} k \paren {k!}^2\) Negated Upper Index of Binomial Coefficient
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} } \sum_{k \mathop = 0}^\infty \dfrac {n! } {k! \paren {k + n}! } \dfrac {\paren {-y + k - 1}!} {k! \paren {-y - 1}!} \dfrac {\paren {-x + k - 1}!} {k! \paren {-x - 1}!} \paren {k!}^2\) Definition of Binomial Coefficient, $\paren {-1}^{2 k} = 1$
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} } \sum_{k \mathop = 0}^\infty \dfrac { \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { k! \paren {n + 1}^{\overline k} }\) Rising Factorial as Quotient of Factorials, and $k!$ cancels


Now expand $\paren {1 + u}^{x + y + n}$ and divide by $u^x$:

\(\ds \dfrac {\paren {1 + u}^{x + y + n} } {u^x}\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \binom {x + y + n} k u^{k - x}\) Binomial Theorem - Complex Numbers


The coefficient $a_n$ of $u^n$ of the product $\dfrac {\paren {1 + u}^{x + y + n} } {u^x}$ above is:

\(\ds a_n\) \(=\) \(\ds \binom {x + y + n} {x + n}\) $n = k - x$, so $k = x + n$
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {x + y + n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + 1} }\) Definition of Binomial Coefficient


Equating coefficients gives us:

\(\ds \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} } \sum_{k \mathop = 0}^\infty \dfrac { \paren {-x}^{\overline k} \paren {-y}^{\overline k} } {\paren {n + 1}^{\overline k} } \dfrac 1 {k!}\) \(=\) \(\ds \dfrac {\map \Gamma {x + y + n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + 1} }\)
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^\infty \dfrac { \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { \paren {n + 1}^{\overline k} } \dfrac 1 {k!}\) \(=\) \(\ds \dfrac {\map \Gamma {x + y + n + 1} \map \Gamma {n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} }\) dividing both sides by $\dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} }$
\(\ds \leadsto \ \ \) \(\ds \map F {-x, -y; n + 1; 1}\) \(=\) \(\ds \dfrac {\map \Gamma {x + y + n + 1} \map \Gamma {n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} }\) Definition of Hypergeometric Function

Letting $a = -x$, $b = -y$ and $c = n + 1$, we obtain:

\(\ds \map F {a, b; c; 1}\) \(=\) \(\ds \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }\)

$\blacksquare$


Source of Name

This entry was named for Carl Friedrich Gauss.


Historical Note

The proof shown above is a more detailed version of a proof by Srinivasa Ramanujan.

Based on Ramanujan's Notebook, as transcribed in Chapter $10$ of Berndt's book, Ramanujan's proof goes as follows:

\(\text {(8.1)}: \quad\) \(\ds \map F {-x, -y; n + 1; 1}\) \(=\) \(\ds \dfrac {\map \Gamma {x + y + n + 1} \map \Gamma {n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} }\)
Assume that $n$ and $x$ are integers with $n \ge 0$ and $n + x \ge 0$.
Expanding $\paren {1 + u}^{y + n}$ and $\paren {1 + \dfrac 1 u}^x$ in their formal binomial series and taking their product, we find that, if $a_n$ is the coefficient of $u^n$:
\(\ds a_n\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \binom {y + n} {k + n} \binom x k\)
\(\text {(8.2)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} } \sum_{k \mathop = 0}^\infty \dfrac { \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { k! \paren {n + 1}^{\overline k} }\)


On the other hand, expanding $\paren {1 + u}^{x + y + n}$ in its binomial series and dividing by $u^x$, we find that:
\(\ds a_n\) \(=\) \(\ds \binom {x + y + n} {x + n}\)
\(\text {(8.3)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {x + y + n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + 1} }\)


Comparing $(8.2)$ and $(8.3)$, we deduce $(8.1)$.


As can be seen, Ramanujan jumped over several intermediate steps in $(8.2)$, but his assertions were all correct.


Sources

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